Question

The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 33 mm and standard deviation 7.2 mm.

(a) What is the probability that defect length is at most 20 mm ?

Answer 1

To find the probability that the defect length is at most 20 mm in a normally distributed dataset with mean 33 mm and standard deviation 7.2 mm, we use the z-score formula and the standard normal distribution table.

Step-by-step explanation:

To find the probability that the defect length is at most 20 mm, we need to calculate the z-score and then use the standard normal distribution table.

The formula for calculating the z-score is: z = (x - μ) / σ, where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

In this case, x = 20 mm, μ = 33 mm, and σ = 7.2 mm. Plugging these values into the formula, we get: z = (20 - 33) / 7.2 = -1.8056.

Using the standard normal distribution table, we can find that the probability corresponding to a z-score of -1.8056 is 0.035. Therefore, the probability that the defect length is at most 20 mm is 0.035 or 3.5%.